Ensino Fundamental ⇒ Triângulo Tópico resolvido

[botelho]

Na figura, calcular RI.( I incentro do triângulo DRO).

IMG-20240423-WA0016.jpg (42.57 KiB) Exibido 115 vezes

a)12
b)13
c)14
d)10
e)9

Resposta

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a

[petras]

botelho,

[tex3]\mathsf{
EI = EO \implies \triangle EIO_{(isos.)} \therefore \angle EIO \cong \angle EOI = 18.5^o \\
\angle OEI=143^o \implies \angle IEG = 37^o \therefore \angle IFG = 2\angle IEG = 74^o \\
\therefore \angle EGI = 90^o \\
\triangle GIE(37^o :53^o:90^o ):(3k:4k:5k) \implies 8k = 4 \therefore k = 2 \implies IE = 2r=10: GI = 6\\
CB \perp RO (B \in RO) \implies \triangle \\

\triangle GIO \cong \triangle BIO \implies BO = 18\\
GA \perp DO (A \in RO) \implies \triangle AGO(37^o:53^o:90^o) \therefore 18 = 4k \implies k = \frac{9}{2}\\
\therefore AO = 5k = \frac{45}{2}\\
AB = AO - BO =\frac{45}{2} - 18 = \frac{9}{2}\\
\triangle ABI(37^o:53^o:90^o) \implies AB = 3k \therefore \frac{9}{2} = 3k \implies k = \frac{3}{2}\\
\therefore IB = 4k = 4.\frac{3}{2}=6\\
\triangle IRB_{(ret)} : sen30^o =\frac{1}{2}=\frac{IB}{RI}=\frac{6}{RI} \therefore \boxed{RI=12}


}

[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
EI = EO \implies \triangle EIO_{(isos.)} \therefore \angle EIO \cong \angle EOI = 18.5^o \\
\angle OEI=143^o \implies \angle IEG = 37^o \therefore \angle IFG = 2\angle IEG = 74^o \\
\therefore \angle EGI = 90^o \\
\triangle GIE(37^o :53^o:90^o ):(3k:4k:5k) \implies 8k = 4 \therefore k = 2 \implies IE = 2r=10: GI = 6\\
CB \perp RO (B \in RO) \implies \triangle \\

\triangle GIO \cong \triangle BIO \implies BO = 18\\
GA \perp DO (A \in RO) \implies \triangle AGO(37^o:53^o:90^o) \therefore 18 = 4k \implies k = \frac{9}{2}\\
\therefore AO = 5k = \frac{45}{2}\\
AB = AO - BO =\frac{45}{2} - 18 = \frac{9}{2}\\
\triangle ABI(37^o:53^o:90^o) \implies AB = 3k \therefore \frac{9}{2} = 3k \implies k = \frac{3}{2}\\
\therefore IB = 4k = 4.\frac{3}{2}=6\\
\triangle IRB_{(ret)} : sen30^o =\frac{1}{2}=\frac{IB}{RI}=\frac{6}{RI} \therefore \boxed{RI=12}


}

[/tex3]

[geobson]

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